# Am I thinking about curve correctly?

#### Movybuf1979

##### New Member
I have been debating with someone on Youtube about what should and shouldn't be visible due to the Earth's curvature. The individual is using Pilot Mountain in North Carolina and saying there is an observatory there that is at 2,200 feet. There are signs that point to other viewable mountains from the observatory. He is pointing to two mountains specifically: Mt. Mitchell and Mt. Rodgers.

Here is what I replied to him:

"Pilot Mountain is 2,421 (He later told me about the observatory at 2,200 feet, but I am sure the difference isn't that great) feet above sea level. Mount Mitchell is 6,684 feet above sea level. They are a little less than 108 miles away from each other. If you plug those numbers into an Earth curve calculator, you will get a hidden height of 1,520.15 feet. This means that you should be able to see 5,163.85 feet of Mount Mitchell from Pilot Mountain."

"Pilot Mountain and Mt. Rogers are about 64 miles apart. As you said the observatory area of Pilot Mt. is at 2,200 feet. Mt. Rogers is 5,729 feet tall. That gives us a Geometric Horizon of 57.44 miles. A geometric drop of 2731.53 feet. A geometric hidden amount of 28.7 feet. Which leaves us with 5,700.3 feet of Mt. Rogers still visible."

He replied to me regarding my numbers for the Pilot Mt/Mt. Mitchel problem with this:

"I don't know who taught you how to do math but you're incorrect, using your number 108 miles, the drop in 108 miles is 7776 Feet (108x108=11664x8"=93312÷12= 7776ft) let's say the land in Between the 2 mountains is 1200 even though it's much greater than that) so you would subtract the difference between 1200 feet and 2200ft (where I was standing)gives you 1000 feet, so you would subtract 1000ft and 6684ft from 7776 which leaves the mountain hidden by 92 feet, so it should be 92 feet below the horizon and you can literally see hundreds of feet when standing there, if not thousands"

I can see that he is not counting the height of the mountains into his evaluation until after accounting for the curve, which doesn't seem like the right way to do it to me. I am not the greatest at math (it's been a long time since I've had to use this kind of math), so I am not sure how to explain to him what/where he is going wrong. Any assistance would be greatly appreciated.

Last edited:

#### Mick West

Staff member
I am not the greatest at math, so I am not sure how to explain to him what/where he is going wrong. Any assistance would be greatly appreciated.
What you really need to do is draw a diagram, and get him to draw a diagram.

#### Movybuf1979

##### New Member
What you really need to do is draw a diagram, and get him to draw a diagram.
When I see the diagram on the Curve calculator on this website the logic makes sense to me. I'm more of a visual learner than a numbers learner. The other person's comments however do not. Unfortunately through YouTube comments, I doubt I would get him to be able to draw a diagram, post it somewhere and link it. Maybe he would, but he seems like the type that doesn't want to listen to reason if it contradicts his personal reality.

Last edited:

#### Mendel

##### Senior Member.
Walter Bislin has a great curve calculator online. For this link, I've preset it with the data for both mountains.
http://walter.bislins.ch/bloge/inde...1572327-32037.2832-11746.1992-4~0.0343-9-21-9 For a more realistic view, go to the "Refraction" tab and click on "Std-Atm". You can also use the slider to match refraction to a specific photograph.

Scrolling further down, you see what the diagram for the first target (Pilot Mountain to Mount Mitchell) looks like.

I have added (in orange) a horizontal reference line and the "drop" that your friend computed. It's easy to see that the drop doesn't actually tell us which part of the mountain should be hidden or not when the observer is at height.

But like Mick says, it's probably better for their understanding if you develop this diagram together.

P.S.: The fact that people can see farther when they are higher up is one of the first BIG clues humanity got that the Earth is round. It's why sailing ships put their observers in the "crow's nest" at the top of the main mast. It's the reason why lighthouses have had their lights at the top of tall towers for millenia (since 280bc).

Last edited:

#### FatPhil

##### Senior Member.
I have been debating with someone on Youtube about what should and shouldn't be visible due to the Earth's curvature. The individual is using Pilot Mountain in North Carolina and saying there is an observatory there that is at 2,200 feet. There are signs that point to other viewable mountains from the observatory. He is pointing to two mountains specifically: Mt. Mitchell and Mt. Rodgers.

Here is what I replied to him:

"Pilot Mountain is 2,421 (He later told me about the observatory at 2,200 feet, but I am sure the difference isn't that great) feet above sea level. Mount Mitchell is 6,684 feet above sea level. They are a little less than 108 miles away from each other. If you plug those numbers into an Earth curve calculator, you will get a hidden height of 1,520.15 feet. This means that you should be able to see 5,163.85 feet of Mount Mitchell from Pilot Mountain."

"Pilot Mountain and Mt. Rogers are about 64 miles apart. As you said the observatory area of Pilot Mt. is at 2,200 feet. Mt. Rogers is 5,729 feet tall. That gives us a Geometric Horizon of 57.44 miles. A geometric drop of 2731.53 feet. A geometric hidden amount of 28.7 feet. Which leaves us with 5,700.3 feet of Mt. Rogers still visible."

He replied to me regarding my numbers for the Pilot Mt/Mt. Mitchel problem with this:

"I don't know who taught you how to do math but you're incorrect, using your number 108 miles, the drop in 108 miles is 7776 Feet (108x108=11664x8"=93312÷12= 7776ft) let's say the land in Between the 2 mountains is 1200 even though it's much greater than that) so you would subtract the difference between 1200 feet and 2200ft (where I was standing)gives you 1000 feet, so you would subtract 1000ft and 6684ft from 7776 which leaves the mountain hidden by 92 feet, so it should be 92 feet below the horizon and you can literally see hundreds of feet when standing there, if not thousands"

I can see that he is not counting the height of the mountains into his evaluation until after accounting for the curve, which doesn't seem like the right way to do it to me. I am not the greatest at math, so I am not sure how to explain to him what/where he is going wrong. Any assistance would be greatly appreciated.

Yup, his maths is bogus, as, as you say, his calculation of drop is for someone standing at the surface, rather than up Pilot Mt. You can't just miss that out it early and then add it back later, that's not how the maths works.

As suggested by our illustrious leader, the best way of understanding the issue is to draw a diagram. Try to be precise, to keep the diagram clear, but you don't need to be accurate - exaggerate the curve of the earth and the height of the peaks.

I'm afraid I can't get my head around your Liberian units, so have converted them to rest-of-the-worldish, and keep angles as radians. Assuming the radius of the earth is 6370km (so somewhere between the equator and the poles), here's a quick verification that an observer at 670m can see an observer at 2037m who is 173.7km away.

? Rearth=6370.
6370.00
? dist=108*1.609
173.772
? angle=dist/Rearth
0.0272797
? H1=2200/3281.
0.670527
? H2=6684/3281.
2.03718
? view1angle=acos(Rearth/(Rearth+H1))
0.0145089
? view1dist=Rearth*view1angle
92.4217
? view2angle=acos(Rearth/(Rearth+H2))
0.0252873
? view2dist=Rearth*view2angle
161.080
? overlap=view1dist+view2dist-dist
79.7298

It's justifiable to ask how that proves my claim at this point, I simply hammered the first few expressions for the most obvious right-angled triangles, and arc angles/lengths. So, in words, the above proves: Mountaineer 1 can see >92km, mountaineer 2 can see >161km and therefore there is a stretch of earth of extent 79.7km that both viewers look down on. From any point along that extent, someone at the surface would be looking up to both mountaineers, and therefore both mountaineers would see each other as being higher than the person on the ground in between.

Hopefully you can see where to put those various distances and angles on your diagrams.

#### Movybuf1979

##### New Member
I'm afraid this is all for my own edification now. I invited the person I was writing to come and defend their position here, but I am very doubtful that they will. As I said this person seems unwilling to accept anything that doesn't fit their conspiracy.

#### Hevach

##### Senior Member.
Unfortunately you'll almost never convince the actual peddlers of misinformation, and rarely the most ardent believers. Challenging them is about reclaiming the casually interested, ensuring fence sitters fall on the right side, and arming the vulnerable against their tricks.

#### Rory

##### Closed Account
"I don't know who taught you how to do math but you're incorrect, using your number 108 miles, the drop in 108 miles is 7776 Feet (108x108=11664x8"=93312÷12= 7776ft) let's say the land in Between the 2 mountains is 1200 even though it's much greater than that) so you would subtract the difference between 1200 feet and 2200ft (where I was standing)gives you 1000 feet, so you would subtract 1000ft and 6684ft from 7776 which leaves the mountain hidden by 92 feet, so it should be 92 feet below the horizon and you can literally see hundreds of feet when standing there, if not thousands"

Yeah, that makes pretty much zero sense. The "drop" figure is irrelevant when talking about hidden amount and "let's say the land in between the two mountains is 1200" (and what follows) seems like something plucked out of thin air.

I guess you'd have to get them to agree on how these things are properly calculated before there could be a sensible discussion. But judging by what they wrote there I wouldn't hold out much hope for that.

I wonder how someone like that would tackle calculating what the hidden amount should be on something they can actually easily measure and verify in the real world? For example: if they're on a flat bit of ground and twenty feet away there's a chair and in between them and the chair there's something that obscures the bottom half of it. Like this:

Clearly drop doesn't come into it, nor does curve or refraction or belief, etc.

Do you think they'd know how to figure out the math for working out how much of the chair they should see?

#### Mendel

##### Senior Member.
Yeah, that makes pretty much zero sense. The "drop" figure is irrelevant when talking about hidden amount and "let's say the land in between the two mountains is 1200" (and what follows) seems like something plucked out of thin air.
No, the fallacy in the thinking has its own logic. It results from thinking about the Globe Earth with Flat Earth concepts (similar to the notion that water would run down the sides of the globe, planes would have to fly "nose down", and others).

"I don't know who taught you how to do math but you're incorrect, using your number 108 miles, the drop in 108 miles is 7776 Feet (108x108=11664x8"=93312÷12= 7776ft) let's say the land in Between the 2 mountains is 1200 even though it's much greater than that) so you would subtract the difference between 1200 feet and 2200ft (where I was standing)gives you 1000 feet, so you would subtract 1000ft and 6684ft from 7776 which leaves the mountain hidden by 92 feet, so it should be 92 feet below the horizon and you can literally see hundreds of feet when standing there, if not thousands"
Content from External Source

It's fairly clear that their understanding of the curvature calculator is "the globe should make mountains drop lower" and they then apply that to a flat sea level concept. Then they add some compensation for the height advantage of the observer ("can see over the 'hump'") in a fashion that still works with their flat-line intuition. It disregards what the globe actually looks like:

Last edited:

#### Mendel

##### Senior Member.
If you can get them to calculate where the "hump" centerpoint is using the drop formula, and then apply straight-line geometry from the observer through centerpoint toward the dropped mountain, that should give an approximately correct result (but does not compensate for atmospheric refraction).